Zeros of Bessel functions

Question

Let $J_\nu(x):=\displaystyle\sum^\infty_{k=0}\frac{(-1)^k(x/2)^{\nu+2k}}{k!~\Gamma(\nu+k+1)}$ denote a Bessel function. When $\nu\geq0$, let $0<j_{\nu,1}<j_{\nu,2}<\cdots$ denote the positive zeroes of $J_\nu(x)$. My questions are:

$(a)$ Keeping $\nu$ fixed, is it known how the $j_{\nu,k}$'s are distributed on the real line (that is, how fast they increase, whether they accumulate somewhere, etc)?

$(b)$ Also, is there a lower or upper bound on the expression $J_{\nu+1}(j_{\nu,k})$ when $\nu$ is held fixed?

Any help is highly appreciated, thanks!

To be honest, I am absolutely stuck on this. I just know the definition of the Bessel function. I have no clue how to make any conclusions about the zeros of a function from an infinite series. The reason I am interested in this is, the terms $j_{\nu,k}$ and $J_{\nu+1}(j_{\nu,k})$ appear in a formula for an $n$-dimensional Bessel process.

Answer 1

I did some googling, and found that a good $($slightly old$)$ reference for this kind of questions is G.N. Watson's “A treatise on the theory of Bessel functions”. In the $1922$ edition that I currently have access to, the relevant theory appears in Chapter XV. See particularly $15.22$, $15.4$ and $15.81$. This does not answer all the questions, of course, but it's a decent amount of information.

In particular, $15.81$ gives an answer to how fast the zeros of $J_\nu(x)$ grow. $15.22$ tells us that the zeros of $J_\nu$ and $J_{\nu + 1}$ are interlaced. But this does not particularly answer question $(b)$ above.

Answer 2

I can help with part (a) of your questions. The distribution of zeroes for the Bessel functions (at least for First Kind, unsure of other ones) on the real line is known. Rather than calculate each zero, I used the scipy module in python to plot the zeroes, from which the pattern can be recognized. Also note that the rate of dampening for fixed nu decreases as x approaches infinity, at which point the spacing between consecutive zeroes approaches pi. For comparison purposes, pi = 3.1415926535897931 (16 decimal digits). From the same python code:

zeros for nu =  0 :
--------------------
[  2.40482556   5.52007811   8.65372791  11.79153444  14.93091771
  18.07106397  21.21163663  24.35247153  27.49347913  30.63460647
  33.77582021  36.91709835  40.05842576  43.19979171  46.34118837
  49.4826099   52.62405184  55.76551076  58.90698393  62.04846919
  65.1899648   68.33146933  71.4729816   74.61450064  77.75602563
  80.89755587  84.03909078  87.18062984  90.32217264  93.46371878]

consecutive differences in zeros: 
- - - - - - - - - - - - - - - - - 
[ 3.11525255  3.1336498   3.13780653  3.13938327  3.14014626  3.14057266
  3.1408349   3.1410076   3.14112734  3.14121375  3.14127814  3.14132741
  3.14136595  3.14139666  3.14142153  3.14144194  3.14145891  3.14147317
  3.14148526  3.14149561  3.14150453  3.14151227  3.14151904  3.14152499
  3.14153024  3.14153491  3.14153907  3.14154279  3.14154614]

zeros for nu =  1 :
--------------------
[  3.83170597   7.01558667  10.17346814  13.32369194  16.47063005
  19.61585851  22.76008438  25.90367209  29.04682853  32.18967991
  35.33230755  38.47476623  41.61709421  44.759319    47.90146089
  51.04353518  54.18555364  57.32752544  60.46945785  63.6113567
  66.75322673  69.89507184  73.03689523  76.17869958  79.32048718
  82.46225991  85.60401944  88.74576714  91.88750425  95.02923181]

consecutive differences in zeros: 
- - - - - - - - - - - - - - - - - 
[ 3.1838807   3.15788147  3.1502238   3.14693811  3.14522846  3.14422587
  3.14358771  3.14315645  3.14285138  3.14262764  3.14245868  3.14232798
  3.14222478  3.14214189  3.1420743   3.14201846  3.1419718   3.14193241
  3.14189885  3.14187004  3.1418451   3.14182339  3.14180436  3.14178759
  3.14177274  3.14175952  3.14174771  3.14173711  3.14172756]

zeros for nu =  2 :
--------------------
[  5.1356223    8.41724414  11.61984117  14.79595178  17.95981949
  21.11699705  24.27011231  27.42057355  30.5692045   33.71651951
  36.86285651  40.00844673  43.15345378  46.29799668  49.44216411
  52.58602351  55.72962705  58.87301577  62.01622236  65.15927319
  68.30218978  71.44498987  74.58768817  77.73029706  80.87282695
  84.01528671  87.15768394  90.30002515  93.44231602  96.58456145]

consecutive differences in zeros: 
- - - - - - - - - - - - - - - - - 
[ 3.28162184  3.20259703  3.17611061  3.16386771  3.15717756  3.15311526
  3.15046124  3.14863095  3.14731501  3.146337    3.14559022  3.14500704
  3.1445429   3.14416743  3.1438594   3.14360355  3.14338872  3.14320659
  3.14305083  3.14291659  3.14280008  3.14269831  3.14260888  3.14252989
  3.14245976  3.14239723  3.14234122  3.14229087  3.14224543]

zeros for nu =  3 :
--------------------
[  6.3801619    9.76102313  13.01520072  16.22346616  19.40941523
  22.58272959  25.7481667   28.90835078  32.06485241  35.21867074
  38.37047243  41.52071967  44.66974312  47.81778569  50.96502991
  54.11161557  57.2576516   60.40322414  63.54840218  66.69324167
  69.83778844  72.9820804   76.12614918  79.27002139  82.41371955
  85.55726287  88.70066784  91.84394868  94.98711773  98.13018573]

consecutive differences in zeros: 
- - - - - - - - - - - - - - - - - 
[ 3.38086123  3.25417759  3.20826544  3.18594907  3.17331437  3.16543711
  3.16018408  3.15650163  3.15381833  3.1518017   3.15024724  3.14902345
  3.14804257  3.14724421  3.14658566  3.14603603  3.14557253  3.14517804
  3.14483949  3.14454677  3.14429196  3.14406878  3.14387221  3.14369816
  3.14354332  3.14340497  3.14328084  3.14316905  3.14306801]

Answer 3

At least in the limit of large $j_{\nu,m}$, one can answer both questions.

From the Digital Library of Mathematical Function, 10.21(iv), we see that for large $m$, $$ j_{\nu,m}\simeq \left(m+\frac\nu2-\frac14\right)\pi+\mathcal O(m^{-1}), $$ which works already quite well for $\nu=0$ already for $m=1$!

Concerning $J_{\mu}(j_{\nu,m})$, one can use the expansion of the Bessel functions for large argument, $$ J_{\mu}(z)\simeq \sqrt{\frac{2}{\pi z}}\cos\left(z-\frac{\mu\pi}2-\frac\pi4\right)+\mathcal O(z^{-1}), $$ to obtain $$ J_{\mu}(j_{\nu,m})\simeq\frac{2 \sqrt{2}}{\pi } \frac{(-1)^m}{\sqrt{4 m+2 \nu -1}} \sin \left(\frac{\pi}{2} ( \nu -\mu)\right) +\mathcal O(m^{-1}). $$